The equation of a circle

93 Topic 4 The equation of a circle Contents 4.1 Revision exercise................................... 94 4.2 The equations of a circle............................... 94 4.2.1 Circles with centre at the origin....................... 94 4.2.2 Circles with centres other than the origin.................. 98 4.2.3 Extended equation of the circle with centre (a, b): a, b 6= 0........ 101 4.3 Circle intersections and tangents.......................... 105 4.4 Summary....................................... 109 4.5 Proofs......................................... 110 4.6 Extended Information................................ 111 4.7 Review exercise................................... 112 4.8 Advanced review exercise.............................. 113 4.9 Set review exercise.................................. 113 Learning Objectives ffl Use the equation of the circle Minimum performance criteria: ffl Given the centre (a, b) and radius r, find the equation of the circle in the form (x - a) 2 +(y-b) 2 =r 2 ffl Find the radius and centre of a circle given the equation in the form x 2 +y 2 + 2gx + 2fy + c = 0 ffl Determine whether a given line is a tangent to a given circle ffl Determine the equation of the tangent to a circle given the point of contact

94 TOPIC 4. THE EQUATION OF A CIRCLE Prerequisites A sound knowledge of the following subjects is required for this unit: ffl Equation of a straight line ffl Distance between two points ffl Algebraic manipulation ffl 2D coordinates and plotting graphs 4.1 Revision exercise Revision exercise 30 min There is a web exercise if you prefer it. Q1: Find the equation of a straight line in the form y=mx+cwith gradient -2 and passing through the point (3, 4) Q2: Find the equation of the straight line perpendicular to the line y = 3x - 4 and passing through the point (0, 1) Q3: Find the distance between the two points (-2, 5) and (3, -4) correct to one decimal place. Q4: Expand and simplify (x - 3) 2 +(y-2) 2 -x 2 + 3(2x - 5) + 4y - 2 Q5: The point P (-3, 4) is reflected in the line y = x to give point Q. Give the coordinates of point Q and the distance between the points P and Q correct to one decimal place. ffl 4.2 The equations of a circle Learning Objective Determine the different forms of the equation of a circle. Find the radius and centre of a circle. ffi Circle A circle is defined as the set of points P (x, y) that are at a given (constant) distance r (the radius) from a given point C (a, b)(the centre). 4.2.1 Circles with centre at the origin fi fl The easiest circle to construct is a circle with centre O, the origin. examples. Here are some

4.2. THE EQUATIONS OF A CIRCLE 95 Equation of a circle with centre, the origin The equation of a circle C with centre O (0, 0) and radius r is x 2 +y 2 =r 2 The proof of this equation is given as proof 1 in the section headed proofs near the end of this topic. Examples 1. What is the radius of a circle with equation x 2 +y 2 =9? Since x 2 +y 2 =r 2 Then r 2 =9 The radius is 3 2. x 2 +y 2 = 25 is the equation of the circle C and the points P and Q are on the circumference with x-coordinate of 4. Find the y-coordinates of P and Q x 2 +y 2 =25 Since P is on the circle, substitute x = 4 into the equation. 16 + y 2 =25) y 2 =9) y=±3

96 TOPIC 4. THE EQUATION OF A CIRCLE 3. What is the radius and centre of the circle with equation 4x 2 +4y 2-36=0? Since this is a circle, rearrange into the correct form. 4x 2 +4y 2 =36) x 2 +y 2 =9 This is in the correct form for a circle with radius 3 units and centre, the origin. 4. Find the equation of a circle passing through the point (5, 12) with centre at the origin. Since the equation is x 2 +y 2 =r 2 Substitute x = 5 and y = 12 into the equation to find r 25 + 144 = r 2 ) 169 = r 2 ) r=13 The equation is x 2 +y 2 = 169 Note that for any circle calculations, the equation must be in the exact form for the equation of a circle with x 2 and y 2 having coefficients equal to one. Circles at the origin exercise 10 min There is a web exercise with randomised parameters if you prefer it. Q6: Find the radii of the following circles with centre, the origin: a) x 2 +y 2 = 144 b) x 2 +y 2-4=0 c) 3x 2 +3y 2-20=7 d) x 2 =-y 2 +8 e) 169 = x 2 +y 2 f) -2x 2 =2y 2-8 Q7: Find the equation of the following circles with centre, the origin: a) The circle with radius 2 b) The circle passing through the point (20, 21) c) The circle with radius 3 p 5 d) The circle passing through the point (4, 0) e) The circle with diameter 10 f) The circle passing through the point on the line y = 2x - 3 where x = 2 Q8: An exercise ring for the stable s horses is circular and located in a square field of side length 20 m. Assuming a clearance of 1 m outside the ring for safety, find the equation of the ring taking its centre to be the origin.

4.2. THE EQUATIONS OF A CIRCLE 97 Q9: A square emblem has a circle at its centre with equation x 2 +y 2 =4cm The pattern repeats to the edge of the emblem with circles on the same centre but with a radius of 1 cm more each time. The emblem side is 10 cm what is the equation of the largest circle on the emblem? Q10: A sweatshirt badge consists of two parallel arrows and a circle as shown. If the centre of the circle is the origin and the arrows touch the circle at the points (-1, 5) and (1, -5) find the equation of the circle.

98 TOPIC 4. THE EQUATION OF A CIRCLE 10 min 4.2.2 Circles with centres other than the origin Calculator activity Using a graphics calculator plot the pairs of values given and investigate the relationship by looking at the centre. On the TI83 calculator, the circles are drawn using 2nd prgm to access the draw menu and then select option 9. a) Circle centre (0, 0) radius 2 and centre (0, 1) radius 2 b) Circle centre (0, 0) radius 4 and centre (-3, 0) radius 4 c) Circle centre (0, 0) radius 5 and centre (2, -3) radius 5 The activity clearly demonstrates that the second circle in each case is a copy of the first circle that has been moved along, down or up to the new centre. Recall that in the functions and graphs topic of this course, the relationships between functions were explored and that certain rules could be applied to find the formula of one function given the other. Here is one example. Example Sketch the graph of y = (x - 3) 2 This is of the form f (x + k) where k = -3 and f (x) = x 2 The effect is a sideways shift of the graph. A rule was given which enabled the graphing of a function related to f (x) and conversely this rule was useful in finding the equation of the related function from a graph. The rule is given below. To obtain y = f (x + k) take y = f (x) and: ffl For k > 0 slide the graph to the left by k units. ffl For k < 0 slide the graph to the right by k units. This rule and the initial activity in this section give an insight into the form of the equation of a circle where the centre is not at the origin.

4.2. THE EQUATIONS OF A CIRCLE 99 General equation of a circle with centre other than the origin The equation of a circle C with centre (a, b) and radius r where a, b 6= 0is (x-a) 2 +(y-b) 2 =r 2 The proof of this equation is given as proof 2 in the section headed proofs near the end of this topic. Thus the circle with equation (x - a) 2 +(y-b) 2 =r 2 can be found by taking the circle with equation x 2 +y 2 =r 2 and For a, b positive: moving it to the right by a units and upwards by b units For a, b negative: moving it to the left by a units and downwards by b units Other variations follow from that rule. Here are some examples. Examples 1. What is the radius and the centre of a circle with equation (x - 2) 2 +(y-5) 2 =9? The radius is 3 and the centre is (2, 5) 2. (x - 1) 2 +(y+3) 2 = 36 is the equation of the circle C and the points P and Q lie on the circumference. If the line through P, Q and C is parallel to the y-axis find the coordinates of P and Q if P lies above Q in a diagram. (x - 1) 2 +(y+3) 2 =36 The centre of the circle is (1, -3) So P and Q have x-coordinate of 1 But the radius is 6 and so P has y-coordinate of -3 +6=3andQhasay-coordinate of -3-6=-9

100 TOPIC 4. THE EQUATION OF A CIRCLE P is the point (1, 3) and Q is the point (1, -9) 3. What is the radius and centre of the circle with equation 3(x + 2) 2 +3(y-1) 2-15= 0? Rearrange into the correct form. 3(x + 2) 2 + 3(y - 1) 2 =15) (x + 2) 2 +(y-1) 2 =5 This is in the correct form for a circle with radius p 5 units and centre with coordinates (-2, 1) 4. Find the equation of a circle passing through the point P (6, 4) with centre at C (-3, 2) The general equation is (x - a) 2 +(y-b) 2 =r 2 The distance between the points P and C is the radius. q (6 + 3) 2 +(4-2) 2 p = 85 = 9.2 The equation is (x + 3) 2 +(y-2) 2 =85 Note that in some instances (see the last example) it is not necessary to calculate the square root as the equation of the circle uses the value of the square of the radius. Points to remember when problem solving: 1. Use the exact formulae for the circle 2. Be aware of the geometry by making a sketch 3. Use the formula for the distance between two points 4. Consider the techniques of straight line equations where appropriate General equation circles exercise 15 min There is an alternative exercise on the web with randomised questions. Q11: Find the radii and centres of the following circles: a) (x + 2) 2 +(y-1) 2 =49 b) x 2 +(y-2) 2-9=0 c) 3(x + 1) 2 + 3(y + 1) 2-2=1 d) (x + 6) 2 =-y 2 +12 e) -6 = -x 2 -(y-1) 2-3 f) -2(x - 1) 2 = 2(y + 4) 2-72 Q12: Find the equation of the following circles: a) The circle with radius 4 and centre (-3, 4) b) The circle passing through the point (4, -5) with centre (2, 2)

4.2. THE EQUATIONS OF A CIRCLE 101 c) The circle with radius 4 p 2 and centre on the positive x-axis such that the origin is on the circumference. d) The circle passing through the point (4, 0) and with centre (-3, -1) e) The circle with diameter 6 and centre midway between the points A (2, 4) and B (-6, 2) f) The circle with diameter PQ where P is the point (8, -4) and Q is the point (2, -12) Q13: A gearing system has two wheels, one large and one small. The line of centres of the two wheels is parallel to the x-axis and the equation of the larger wheel is (x - 2) 2 +(y+4) 2 = 64. Find the equation of the smaller wheel which lies to the right of the larger wheel and has a radius of half of that of the larger wheel. Q14: A child s bicycle has two identical wheels with a clearance between them of 6 inches. When held against a wall (represented by the y-axis with the ground as the x- axis) the bike measures 4 1 / 2 feet lengthwise. Find the equations of the two circles which represent the wheels. 4.2.3 Extended equation of the circle with centre (a, b): a, b 6= 0 The general form of the equation of a circle can be expanded algebraically.

102 TOPIC 4. THE EQUATION OF A CIRCLE (x-a) 2 +(y-b) 2 =r 2 ) x 2-2ax+a 2 +y 2-2by+b 2 -r 2 =0) x 2 +y 2 + 2gx + 2fy + c = 0 where g = -a, f = -b and c = a 2 +b 2 -r 2 This form avoids confusion with a and b by using g and f and also uses addition in the equation instead of subtraction. Extended equation of the circle The extended equation of the circle where (-g, -f) is the centre and p (g 2 +f 2 - c) is the radius (provided g 2 +f 2 -c> 0) is x 2 +y 2 + 2gx + 2fy + c = 0 The derivation is also included in the section on proofs for completeness as proof 3. Examples 1. Find the centre and radius of the circle represented by the equation x 2 +y 2 +10x+6y-2=0 This is in the extended form of the circle x 2 +y 2 + 2gx + 2fy + c = 0 where g = 5 and f = 3 The centre is (-5, -3) r 2 =25+9+2=36sor=6 The radius is 6 2. In extended form give the equation of the circle with radius of 4 and centre (-6, 2) The centre is represented by (-g, -f) ) g=6andf=-2 r 2 =g 2 +f 2 -c) c = 36 + 4-16 = 24 The equation is x 2 +y 2 +12x-4y+24=0 3. The centres of two grinding wheels have to be at least 18 cm apart for safety. The equation of one of the wheels is x 2 +y 2 + 4x + 2y - 4 = 0. If the clearance between the wheels must be at least 1 cm find the range of values which the radius of the second wheel can take.

4.2. THE EQUATIONS OF A CIRCLE 103 If the clearance is set at 11 cm find the equation of the second wheel given that the centres of the two wheels lie on a horizontal bench. From the equation g = 2, f = 1 and r 2 =g 2 +f 2 -c=9) the radius is 3 cm With a clearance of at least 1 cm the edge of the second wheel is at least 4 cm away; but the centres are at least 18 cm apart ) the second wheel radius must be at the most 14 cm The second wheel radius has a range of 0 < r» 14 If the clearance is then set at 11 cm, the second wheel has a radius of 4 cm The x-coordinate of the centre will be -2 + 18 = 16 The y-coordinate is the same for both wheels at y = -1 The equation for the second wheel is x 2 +y 2-32x + 2y + 241 = 0 4. Find the equation of the circle through the three points A (-2, 1), B (1, 4) and D (-2, 7) (This is an alternative way to some textbooks) Let the centre of the circle be C (a, b), then AC = BC = DC = radius Using the distance between two points gives AC 2 =(a+2) 2 +(b-1) 2 AB 2 =(a-1) 2 +(b-4) 2 DC 2 =(a+2) 2 +(b-7) 2 Equate AC 2 with AB 2 and simplify to give the equation b = -a + 2: call this equation 1 Equate AC 2 with DC 2 and simplify to give the equation b = 4: call this equation 2 Solving equations 1 and 2 gives a = -2 and b = 4 The centre is (-2, 4) and the radius is found by substituting, say in AC 2, to give AC 2 = 9. That is r = 3. The equation is x 2 +y 2 +4x-8y+11=0

104 TOPIC 4. THE EQUATION OF A CIRCLE The last example demonstrates that a good grounding in geometry will help with the work on circles. In fact this last example could easily be solved in other ways. For example, the fact that perpendicular bisectors of chords pass through the centre of a circle could be used. The equations of the perpendicular bisectors of the two chords AB and BD can be equated to give the centre of the circle. Geometry in circles investigation 15 min Take the last example which was: Find the equation of the circle through the three points A (-2, 1), B (1, 4) and D (-2, 7) Use the perpendicular bisector method to check the result. What other geometric facts can be ascertained about these three points? Does this suggest another approach to solving this problem? Extended equation of a circle exercise 15 min There is a web exercise if you prefer it. Q15: Find the radii and centres of the following circles: a) The circle with equation x 2 +y 2-4x-2y-4=0 b) The circle with equation x 2 +6x-1=-y 2 +4y+2 c) The circle with equation x 2 +y 2 +8y-9=0 d) The circle with equation x 2 +y 2-10x+21=0 Q16: Find the equation in extended form of the following circles: a) The circle with radius 3 and centre (-2, 1) b) The circle with diameter 8 and centre (0, 5) c) The circle with diameter through the points (2, 3) and (8, -3) d) The circle C with its centre on the circumference of the circle A at x=-2andwith the same radius. Circle A has equation x 2 +y 2-10x+8y-8=0 Q17: A company logo consists of three circles with centres at each vertex of a right angled triangle with sides of 3 cm, 4 cm and 5 cm as shown. Circle A has equation x 2 +y 2-2x+4y+4=0. Allcircles are only touching (not intersecting). Find the equations of the remaining circles A and C. The logo is shown as it hangs with AB vertical. What is the radius and centre of circle D?

4.3. CIRCLE INTERSECTIONS AND TANGENTS 105 ffl 4.3 Circle intersections and tangents Learning Objective ffi Solve problems including intersections with circles and lines The problems considered previously were based solely on circles and the relationship with other circles. Many problems however, are concerned with the relationship between lines and circles. Geometrically, a circle and line can: 1. not intersect 2. intersect at two distinct points 3. touch at one point (in effect this is intersection at two equal points) fi fl When a line meets a circle at one point, it is called a tangent to the circle. Tangents and circles have many geometric properties, some of which will be used in this section. Examples 1. Intersection of a line and a circle Find where the line y = -2x - 3 meets the circle with equation x 2 +y 2-4x+2y-4=0 Substitute the value y = -2x - 3 into the equation of the circle to give x 2 + (-2x - 3) 2-4x + 2(-2x - 3) -4=0 x 2 +4x 2 +12x+9-4x-4x-6-4=0 5x 2 +4x-1=0 (5x - 1)(x + 1) = 0 ) x= 1 / 5 orx=-1 When x = 1 / 5,y=-2 1 / 5-3= -17 / 5 When x = -1, y = -2-1-3=-1 The two points are ( 1 / 5, -17 / 5 ) and (-1, -1) 2. Tangent to a circle Find the equation of the tangent to the circle x 2 +y 2 = 25 at the point P (3, 4)

106 TOPIC 4. THE EQUATION OF A CIRCLE The circle centre is the O, the origin. The gradient of OP = 4 / 3 The geometry of tangent shows that the gradient of a tangent at a point is at right angles to the gradient of the radius to that point. Therefore the gradient of the tangent at P is -3 / 4 The tangent is a straight line. Thus the equation of the tangent is (y - 4) = -3 / 4 (x-3) That is 4y = -3x + 25 3. Does the line y = 2x + 5 intersect the circle with equation (x - 2) 2 +(y+1) 2 =4? Substitute y = 2x + 5 into the equation of the circle. x 2-4x+4+(2x+6) 2 =4 x 2-4x+4+4x 2 +24x+36-4=0 5x 2 +20x+36=0 Look at the discriminant D ( = b 2-4ac). D = 400-720 = -320 Since the discriminant is negative there is no solution and the line and circle do not meet. 4. Intersection of circle and tangent Find the point at which the circle with equation x 2 +y 2 +6x-8y-7=0andthetangent y = x - 1 meet. Substitute y = x - 1 into the equation of the circle to give x 2 +x 2-2x+1+6x-8x+8-7=0 2x 2-4x+2=0 (2x - 2)(x - 1) = 0 ) x = 1 twice. (therefore it is a tangent!). When x = 1, y = x - 1 = 0 The point at which the circle and tangent meet is (1, 0)

4.3. CIRCLE INTERSECTIONS AND TANGENTS 107 5. Find the two values of k such that the line y = -2x + k is a tangent to the circle x 2 +y 2-8x-2y+12=0 Substitute for y in the equation of the circle to give x 2 + (-2x + k) 2-8x - 2(-2x + k) + 12 = 0 5x 2 + x(-4-4k) + k 2-2k+12=0 For the line to be a tangent, the discriminant is zero (-4-4k) 2-4 5 (k 2-2k+12)=0-4k 2 + 72k - 224 = 0 k 2-18k+56=0 (k - 4)(k - 14) = 0 ) k=4ork=14 It is also possible to determine if two circles touch or intersect. Again much of the technique is dictated by the particular problem. The next example relies on the distance formula. Example Determine if the two circles with equations (x - 2) 2 +(y-3) 2 = 16 and (x + 1) 2 +(y-3) 2 = 25 touch, intersect or avoid each other. This is one way in which it may be possible to decide without reverting to looking for intersection points. Let C be the circle with equation (x - 2) 2 +(y-3) 2 =16 and D be the circle with equation (x + 1) 2 +(y-2) 2 =25 C has centre (2, 3) and radius 4 and D has centre (-1, 2) and radius 5 The distance between the centres is found by using the distance formula and is p 10 = 3.2 to 1 d.p. The two radii are both greater than this distance and so the circles intersect. Beware: there are times where one circle could lie completely within another circle and so does not intersect. Using the approach shown in the last example works in some cases only. It will not work if one circle has a radius less than the distance between the

108 TOPIC 4. THE EQUATION OF A CIRCLE centres when the second circle has a radius greater than this distance plus the radius of the smaller circle. Here is an example. Circles within circles 20 min Investigate circles within circles by varying the distance between the centres for different radii. Try to confirm the statement given in the last paragraph, viz, circles do not intersect if one circle has a radius less than the distance between the centres when the second circle has a radius greater than this distance plus the radius of the smaller circle. Find another condition when they do not intersect. Returning to the example the following method will provide the coordinates of the points of intersection of the two circles. Example : Intersecting circles Find the intersection points of the two circles with equations (x-2) 2 +(y-3) 2 = 16 and (x + 1) 2 +(y-3) 2 =25 The intersection points can be found by solving the two equations simultaneously. (x-2) 2 +(y-3) 2-16 = 0: equation 1 (x+1) 2 +(y-3) 2-25 = 0: equation 2 equation 1 - equation 2 gives (x-2) 2-16 - (x + 1) 2 +25=0 x 2-4x+4-16-x 2-2x-1+25=0-6x+12=0) x=2 Substitute in one equation, say, (x - 2) 2 +(y-3) 2 =16togive: (y-3) 2-16=0) y 2-6y-7=0) (y - 7)(y + 1) = 0 y=7ory=-1 The two points of intersection are (2, 7) and (2, -1) If the two circles touch instead of intersect, the quadratic equation will give two equal values on substitution.

4.4. SUMMARY 109 Q18: What condition will ensure that two circles touch but do not intersect? This can in fact be used to show that circles touch. Intersection exercise There is a web exercise if you prefer it. Q19: When y = 0, there are two tangents to the circle x 2 +y 2 +6x-8y-7=0.Find the equations of these tangents. 30 min Q20: Find the equation of the tangent to the circle x 2 +y 2-2x + 4y - 4 = 0 at the point P (2, -5) Q21: Find the points of intersection of the line y = 2x - 4 and the circle with equation x 2 +y 2-5x-2y-54=0 Q22: Determine whether the line y = x + 3 is a tangent to the circle with equation x 2 +y 2 + 4x + 2y + 3 = 0. Consider the line y = -x + 3: does this line touch, intersect or avoid the circle? Q23: Find the intersection points of the circles with equations x 2 +y 2 +4x-6y-3=0 and x 2 +y 2-2x-6y-9=0 Q24: Find k such that the line y = -x + k is a tangent to the circle with equation x 2 +y 2-10x-2y+18=0 Q25: Find k such that the line y = -2x + k is a tangent to the circle with equation x 2 +y 2-4x-4y+3=0 4.4 Summary The following points and techniques should be familiar after studying this topic: ffl The equation of the circle in the form (x - a) 2 +(y-b) 2 =r 2. ffl The equation in the form x 2 +y 2 + 2gx + 2fy + c = 0. ffl The points of intersection of lines and circles or two circles. ffl The equations of tangents to a circle and points of contact of tangents.

110 TOPIC 4. THE EQUATION OF A CIRCLE 4.5 Proofs Proof 1: x 2 +y 2 = r is a circle centre the origin. Let P be any point (x, y) on the circumference of the circle. Then OP = radius, r By the formula for the distance between two points, q r= (x-0) 2 +(y-0) 2 = p x 2 +y 2 It follows that x 2 +y 2 =r 2 Proof 2: (x - a) 2 +(y-b) 2 =r 2 is a circle centre (a, b) and radius r Let the circle have a centre at the point C (a, b) for any a, b 6= 0 Let P be any point (x, y) on the circumference of the circle. By definition, r is the radius and is the distance CP

4.6. EXTENDED INFORMATION 111 but by the distance formula, the length of CP is q (x - a) 2 +(y-b) 2 Thus (x - a) 2 +(y-b) 2 =r 2 Proof 3: x 2 +y 2 + 2gx + 2fy + c = 0 is a circle with centre (-g, -f) and radius p (g 2 +f 2 -c) This is not a proof. It is a derivation of another formula from the one shown at proof 2. The equation (x - a) 2 +(y-b) 2 =r 2 represents a circle with centre (a, b) and radius r (x - a) 2 +(y-b) 2 =r 2 ) x 2-2ax+a 2 +y 2-2by+b 2 -r 2 =0 Let -g = a, -f = b then the centre of the circle (a, b) is expressed as (-g, -f) x 2 +2gx+g 2 +y 2 + 2fy + f 2 -r 2 =0 Letc=g 2 +f 2 -r 2 Then the equation becomes x 2 +2gx+y 2 +2fy+c=0 Rearranging c = g 2 +f 2 -r 2 gives r = p (g 2 +f 2 - c) as required. ffl 4.6 Extended Information Learning Objective ffi Find resources on the web for circle study There are links on the web which give a selection of interesting sites to visit. Browsing the web under circle will lead to many other good sites which cover this topic. The study of circles is one of the more ancient topics with discoveries going back into the distant centuries. Many of the properties were discovered by the Egyptians and Greeks. The history of the Mathematicians who studied circles is naturally complex and combined with studies of curves and other geometric shapes. The mathematicians detailed here give an insight into the vast amount of work and study undertaken. In this topic it is best for the reader to search for topics which are of interest, such as squaring the circle, investigations on spirals and ellipses. On the web, the St Andrews University site provides an excellent link to some of the Mathematicians. Euclid Euclid solved all his problems using logical reasoning and concentrated on the geometry of circles and straight lines. Thales This Greek mathematician formulated some of the earliest theorems on the circle in the seventh century BC. fi fl

112 TOPIC 4. THE EQUATION OF A CIRCLE Hippocrates One of the most famous investigations is that of squaring a circle. Hippocrates spent much of his life working on this problem. Ramanujan Mathematicians through the centuries continue to wrestle with the problem but Ramanujan gave a construction in the early 1900s which is accurate to 6 d.p. Plato Plato was a great astronomer who believed that the planets were spherical. He was fascinated by geometric constructions and much of his work laid the foundations for others to follow. 4.7 Review exercise Review exercise 30 min There is another exercise on the web if you prefer it. Q26: A circle has radius 3 units and centre (-4, -2). Write down the equation of the circle in general form. Q27: A circle has equation x 2 +y 2-2x + 4y - 11 = 0. Write down the coordinates of the centre of the circle and the length of the radius. Q28: Show that the straight line with equation y = 2x +1isatangent to the circle with equation x 2 +y 2-10x-12y+56=0 Q29: The point P (-1, 2) lies on a circle with centre C (1, -2). Find the equation of the circle and of the tangent at P Q30: A circle has equation x 2 +y 2-8x + 2y - 8 = 0. Write down the coordinates of the centre of the circle and the length of the radius. Q31: The point P (3, -2) lies on a circle with centre C (9, -5). Find the equation of the tangent at P

4.8. ADVANCED REVIEW EXERCISE 113 4.8 Advanced review exercise Advanced review exercise There is another exercise on the web if you prefer it. Q32: A new company sign has a logo in the shape of two circles, one sitting on top of the other as shown and 22 cm high. 30 min The equation of the smaller circle is x 2 +y 2-12x - 26y + 189 = 0 and the line of centres is parallel to the y-axis. Find the equation of the larger circle. Q33: The company decides that the logo is better with the large circle on the bottom. If the equation of the smaller circle remains the same, what is the new equation of the larger circle? Q34: Three circular cogs in a machine are set so that the largest (and central) cog A has equation x 2 +y 2-2x+4y-4=0.Thetwosmaller cogs B and C have radii of 1 and 2 units respectively and are set so that AB is horizontal and BC is perpendicular to AB. Cog B touches cog A and cog C. If the cog B lies to the right of cog A and C lies above B, find the equations of the two circles which represent cogs B and C in the general form of (x - a) 2 +(y-b) 2 =r 2 and state with reasons whether cogs A and C touch, intersect or avoid contact with each other. Q35: The circles A and B with centres at (0, -1) and (-1, -1) intersect. If the radius of A is p 40 and the radius of B is p 45, find the point (with positive coefficients) at which the circles cross. Give the equation of the lines through each centre to this point. 4.9 Set review exercise Set review exercise Work out your answers and then access the web to input your answers in the online test called set review exercise. 30 min Q36: A circle has radius 5 units and centre (-2, 4). Write down the equation of the circle.

114 TOPIC 4. THE EQUATION OF A CIRCLE Q37: A circle has equation x 2 +y 2 + 8x - 2y - 8 = 0. Write down the coordinates of the centre of the circle and the length of the radius. Q38: Show that the straight line with equation y = 2x -3isatangent to the circle with equation x 2 +y 2-2x-8y+12=0 Q39: A circle has equation x 2 +y 2-2x + 6y - 6 = 0. Write down the coordinates of the centre of the circle and the length of the radius. Q40: The point P (-4, -1) lies on a circle with centre C (-2, 2). Find the equation of the tangent at P Q41: The point P (-1, 2) lies on a circle with centre C (3, 4). Find the equation of the tangent at P

ANSWERS: TOPIC 4 149 4 The equation of a circle Revision exercise (page 94) Q1: Use the formula (y - y 1 ) = m(x - x 1 ) to give y-4=-2(x-3) y=-2x+10 Q2: Let L1 be y = 3x - 4 and L2 be the line perpendicular to it and passing through the point (0, 1) Perpendicular lines have gradients m 1 and m 2 such that m 1 m 2 =-1 The gradient of the line L2 = - 1 / 3 Using the formula gives y - 1 = - 1 / 3 (x-0) y=- 1 / 3 x+1or3y=-x+3 q Q3: Use the formula (x - x 1 ) 2 +(y-y 1 ) 2 The distance is: q (3+2) 2 +(-4-5) 2 = p 25 + 81 = 10.3 Q4: (x-3) 2 +(y-2) 2 -x 2 + 3(2x - 5) + 4y - 2 = x 2-6x+9+y 2-4y+4-x 2 +6x-15+4y-2= y 2-4 or (y - 2)(y + 2) Q5: Draw it and see. On reflection the positive y-axis becomes the positive x-axis and the negative also switch. The point Q is therefore (4, -3) The distance between P and Q is p [(4+3) 2 (-3-4) 2 ] = 9.9 units correct to one decimal place. Circles at the origin exercise (page 96) Q6: a) Radius is 12 b) Radius is 2 c) x 2 +y 2 =9) radius is 3 d) Radius is p 8or2 p 2 e) Radius is 13 f) x 2 +y 2 =4) radius is 2 Q7: a) x 2 +y 2 =4 b) 20 2 +21 2 =r 2 so r = 29 The equation is x 2 +y 2 = 841 c) x 2 +y 2 =45

150 ANSWERS: TOPIC 4 d) x 2 +y 2 =16 e) x 2 +y 2 =25 f) If x = 2, y = 1 and 2 2 +1 2 =r 2 so r = p 5 The equation is x 2 +y 2 =5 Q8: The diameter of the ring is 18 m and the radius is 9 m The equation is x 2 +y 2 =81 Q9: The radius of the first circle is 2 cm which is a diameter of 4 cm. The next circle has diameter of 6 cm, then 8 cm and so the largest circle has diameter of 10 cm and therefore a radius of 5 cm The equation of the largest circle is x 2 +y 2 =25 Q10: The length between the two points where the arrows touch the centre is the diameter. This length is calculated with the distance formula. q r= (1 + 1) 2 +(-5-5) 2 = p 104 The equation of the circle is x 2 +y 2 = 104 General equation circles exercise (page 100) Q11: a) Radius is 7 and the centre is (-2, 1) b) Note that x 2 is the same as (x - 0) 2. Rearrange into the general form. Radius is 3 and the centre is (0, 2) c) (x + 1) 2 +(y+1) 2 =1) radius is 1 and the centre is (-1, -1) d) Radius is p 12 or 2 p 3 and the centre is (-6, 0) e) Rearrange to give x 2 +(y-1) 2 =3 Radius is p 3 and the centre is (0, 1) f) Rearrange to give (x - 1) 2 +(y+4) 2 =36 The radius is 6 and the centre is (1, -4) Q12: a) (x + 3) 2 +(y-4) 2 =16 b) The radius r is the distance between the centre and the point. so r 2 =(4-2) 2 + (-5-2) 2 =53 The equation is (x - 2) 2 +(y-2) 2 =53 c) The centre is (4 p 2, 0) The equation is (x - 4 p 2) 2 +y 2 =32 d) r 2 =(4+3) 2 +(0+1) 2 =50 The equation is (x + 3) 2 +(y+1) 2 =50 e) Radius is 3 The midway point is (-2, 3) The equation is (x + 2) 2 +(y-3) 2 =9

ANSWERS: TOPIC 4 151 f) The distance between P and Q is q (8-2) 2 +(-4+12) 2 = p 100 = 10 The radius is 5 The centre of the circle is the midway point between P and Q which is ( 5, -8) The equation is (x - 5) 2 +(y+8) 2 =25 Q13: The centre of the larger wheel is (2, -4) and the radius is 8 Thus the smaller wheel has radius 4 The centre of the smaller wheel is 12 units to the right of the larger wheel. The centre is at (14, -4) The equation of the smaller wheel is (x - 14) 2 +(y+4) 2 =16 Q14: The two wheels measure 4 feet so each is 2 feet across and therefore has a radius of 1 foot. The first wheel A has centre at (1, 1) since it rests against the wall and the ground. It has equation (x - 1) 2 +(y-1) 2 =1 The second wheel centre is 2 1 / 2 feet away and has a centre at (3.5, 1) It has equation (x - 3.5) 2 +(y-1) 2 =1 Extended equation of a circle exercise (page 104) Q15: a) The centre is given by (2, 1) and the radius is 3 b) The centre is given by (-3, 2) and the radius is 4 c) The centre is given by (0, -4) and the radius is 5 d) The centre is given by (5, 0) and the radius is 2 Q16: a) x 2 +y 2 +4x-2y-4=0 b) x 2 +y 2-10y+9=0 c) x 2 +y 2-10x+7=0 d) x 2 +y 2 +4x+8y-29=0 Q17: Circle A has centre (1, -2) and radius 1. Some geometry is needed. As AB = 3 and circles A and B touch then circle B has radius 2 Circle B has centre with x-coordinate of 1 (as in A) and y -coordinate of -2-1(radius of A) - 2 (radius of B) = -5 So for circle B with r = 2, g=-1andf=5theequation is x 2 +y 2-2x+10y+22=0 Since BC has length 4 and circle B has radius of 2, circle C has also radius of 2 The centre of C has y -coordinate of -5 (as has B) and x-coordinate of 1 + 2 + 2 = 5

152 ANSWERS: TOPIC 4 The equation of circle C is x 2 +y 2-10x+10y+46=0 Since circle A has radius 1 and circle C has radius 2, then circle D has diameter 2 since AC = 5 It follows that circle D has radius of 1 and the centre lies along AC at a point in the ratio 2:3 The centre is (1 + 8 / 5,-2-6 / 5 )=( 13 / 5, -16 / 5 ) Answers from page 109. Q18: The sum of the two radii of the circles is equal to the distance between the centres. Intersection exercise (page 109) Q19: When y = 0 the equation becomes x 2 +6x-7=0 (x + 7)(x - 1) = 0 ) x = -7 or x = 1 The points are P (1, 0) and Q (-7, 0) The centre of the circle is C (-3, 4) m CP = -1 therefore m tan = 1 at this point P Using the equation of a straight line (y - b) = m(x - a) gives y=x-1 m CQ =1som tan =-1 In this case y = - x - 7 Q20: The circle has centre C (1, -2) m CP = -3 / 1 =-3 m tan = 1 / 3 The equation is given by (y - a) = m(x - a) y+5= 1 / 3 (x-2) 3y=x-17 Q21: Substitute y = 2x - 4 into the equation of the circle. x 2 +4x 2-16x+16-5x-4x+8-54=0 5x 2-25x - 30 = 0 x 2-5x-6=0 (x - 6)(x + 1) = 0 ) x=6orx=-1 When x = 6, y = 8 giving the point (6, 8) When x = -1, y = -6 giving the point (-1, -6)

ANSWERS: TOPIC 4 153 Q22: Substitute y = x + 3 into the equation of the circle. x 2 +x 2 +6x+9+4x+2x+6+3=0 2x 2 + 12x + 18 = 0 x 2 +6x+9=0 EITHER (x + 3)(x + 3) = 0 ) that the line is a tangent as the two values are equal OR the discriminant b 2-4ac = 36-36 = 0 ) the roots are equal ) the line is a tangent. Substituting y = -x + 3 into the circle equation and simplifying gives x 2-2x+9=0 But the discriminant =4-36=-32) no real solutions ) the line y = -x + 3 does not come into contact with the circle. Q23: General or extended form can be used. Here is general form. x 2 +y 2 + 4x - 6y - 3 = 0 rearranges to give (x + 2) 2 +(y-3) 3 =16 x 2 +y 2-2x - 6y - 15 = 0 rearranges to give (x - 1) 2 +(y-3) 2 =25 At intersection these solve simultaneously to give (x + 2) 2-16-(x-1) 2 +25=0 x 2 +4x+4-16-x 2 +2x-1+25=0 6x + 12 = 0 ) x=-2 Substitute x = -2 into say, x 2 +y 2 + 4x - 6y - 3 = 0 to give 4+y 2-8-6y-3=0 (y - 7)(y + 1) = 0 ) y = 7 or y = -1 and the two intersection points are (-2, 7) and (-2, -1) Q24: Substitute for y in the circle to give x 2 +(-x+k) 2-10x - 2(-x + k) + 18 = 0 This simplifies to 2x 2 + x(-8-2k) + k 2-2k+18=0 The discriminant gives k 2-12k + 20 = 0 for the line to be a tangent. k=2ork=10 Q25: Substitute for y in the circle to give x 2 + (-2x + k) 2-4x - 4(-2x + k) + 3 = 0 This simplifies to 5x 2 + x(4-4k) + k 2-4k+3=0 The discriminant gives k 2-12k+11=0 k=1ork=11 Review exercise (page 112) Q26: The equation takes the form (x - a) 2 +(y-b) 2 =r 2 where (a, b) is the centre of the circle and r is the radius. The equation of this circle is (x + 4) 2 +(y+2) 2 =9

154 ANSWERS: TOPIC 4 Q27: x 2 +y 2 + 2gx + 2fy + c = 0 is a general equation of a circle This circle has g = -1, f = 2 and c = -11 The centre is given by (-g, -f) and thus the centre is (1, -2) r 2 =g 2 +f 2 -c=1+4+11=16 The radius is 4 Q28: Substitute y = 2x + 1 into the equation of the circle to give: x 2 +4x 2 + 4x + 1-10x - 24x - 12 + 56 = 0 ) 5x 2-30x+45=0) x 2-6x+9=0) (x-3) 2 =0) x = 3 twice ) it is a tangent and touches at this one point. Q29: The circle has centre (1, -2) r 2 =4+16=20 The equation of the circle is (x-1) 2 +(y+2) 2 =20 or alternatively x 2 +y 2-2x+4y-15=0 The gradient of PC is -2 ) the gradient of the tangent is 1 / 2 y-2= 1 / 2 (x+1) The equation of the tangent is 2y = x + 5 Q30: The centre is (4, -1) and the radius is 5 Q31: The gradient of PC is - 1 / 2 ) gradient of the tangent is 2 Thusy+2=2(x-3)) the equation of the tangent is y = 2x - 8 Advanced review exercise (page 113) Q32: Let the small circle be circle S and the large circle be circle L The centre of circle S is (6, 13) and the line of centres has equation x = 6 since it is parallel to the y-axis. The x-coordinate of the centre of circle L is 6 The radius of circle S = p (6 2 +13 2-189) = 4 Therefore the diameter of circle L is 22-8 cm = 14 cm and the radius is 7 cm The distance from circle S centre to circle L centre is 11 cm and so the y-coordinate of the centre of circle L is 24 Thus circle L has centre (6, 24) and radius 7 cm The equation is (x - 6) 2 + (y - 24) 2 =49or x 2 +y 2-12x - 48y - 563 = 0 Q33: The distance from circle S centre to circle L centre is still 11 cm and the y -coordinate is 13-11 = 2 The circle has centre (6, 2) and radius 7 cm

ANSWERS: TOPIC 4 155 The new equation is (x - 6) 2 +(y-2) 2 =49or x 2 +y 2-12x - 4y - 9 = 0 Q34: Cog A has centre (1, -2) and radius 3 Cog B has radius 1 and centre with x-coordinate of 1 + 3+1=5 Thus it has centre (5, -2) and the equation of cog B is (x - 5) 2 +(y+2) 2 =1 Cog C has centre at a y-coordinate of -2 +1+2=1andxcoordinate of 5 Since its radius is 2 units, the equation of cog C is (x - 5) 2 +(y-1) 2 =4 Use the distance formula to give AC =5cm But radius A + radius B=5cm Therefore the two circles touch. Alternatively a sketch will reveal that the centres form a right angled triangle and that cogs A and C do in fact touch. Q35: Circle A has equation x 2 +(y+1) 2 = 40 and Circle B has equation (x + 1) 2 +(y+1) 2 =45 At a common point the equations will solve simultaneously to give x 2-40-(x+1) 2 +45=0) 2x = 4 x=2 Substitution in circle A gives 4 + (y + 1) 2 =40) (y + 7)(y - 5) = 0 ) y = 5 since the tangent point required has positive coefficients. The common tangent is at the point (2, 5)

156 ANSWERS: TOPIC 4 For circle A the gradient is 3 ) y-5=3(x-2)) y=3x-1 For circle B the gradient is 2 ) y-5=2(x-2)) y=2x+1 Set review exercise (page 113) Q36: This answer is only available on the web. Q37: This answer is only available on the web. Q38: This answer is only available on the web. Q39: This answer is only available on the web. Q40: This answer is only available on the web. Q41: This answer is only available on the web.